The answer is: #1/16(pi^2+4)#
The first thing to do is to lower the degree of the function sinus with the Half-angle formula:
#sinx=sqrt((1-cos2x)/2#.
So the integral becomes:
#int_0^(pi/2)x(1-cos2x)/2dx=1/2int_0^(pi/2)(x-xcos2x)dx=1/2[int_0^(pi/2)xdx-int_0^(pi/2)xcos2xdx]#.
To do the first integral it is sufficient to remember the law:
#intx^ndx=x^(n+1)/(n+1)+c#,
so:
#int_0^(pi/2)xdx=[x^2/2]_0^(pi/2)=[(pi^2/4)/2-0/2]=pi^2/8#
The second integral can be done using the theorem of the integration by parts, that says:
#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#
We can assume that #f(x)=x# and #g'(x)dx=cos2xdx#.
To find #g(x)#, it is sufficient to remember this law:
#g(x)=intcos(h(x))h'(x)dx=sin(h(x))+c#,
where #h(x)=2x, and h'(x)=2#,
So:
#g(x)=intcos2xdx=1/2intcos2x.2dx=1/2sin2x#.
and #f'(x)=1#
So:
#int_0^(pi/2)xcos2xdx=[x*1/2sin2x]_0^(pi/2)-int_0^(pi/2)1*1/2sin2xdx=#
#[pi/2*1/2sin2(pi/2)-0]-1/2[-(cos2x)/2]_0^(pi/2)=[pi/4sinpi+1/2(cos2(pi/2))/2-1/2cos0/2]=[0-1/4-1/4]=-1/2.#
I remember that #intsin2xdx=-(cos2x)/2#, done like before with #intcos2xdx=(sin2x)/2#
Final count:
#1/2(pi^2/8+1/2)=1/16(pi^2+4)#
